bzoj1875 HH去散步

HH去散步

题目背景：

bzoj1875

分析：矩阵快速幂优化DP

 

这道题有了不能从一条边(x, y)在相邻时刻从x à y，然后y à x，那么本来我想的方式是，定义点对(j, k)上一步是从j à k，那么这一步就只能更新除了从k à j以外的位置，然后发现题目当中有说可能有重边，然后我就想考虑将重边的点重新copy一个，然后就发现这样的复杂度直接就爆掉了······最后发现原来这道题是一道妙妙的用矩乘维护边的转移，我们直接将双向边拆成单项，然后从0标号，那么正向边和反向边的标号x, y就是x = y ^ 1, y = x ^ 1，那么每次转移的时候除了不能用反向边，其他的都可以，直接枚举就可以了。

Source:

/*
	created by scarlyw
*/
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cctype>
#include <vector>
#include <set>
#include <queue>

const int MAXN = 120 + 10;
const int mod = 45989;

struct node {
	int to, id;
	node(int to = 0, int id = 0) : to(to), id(id) {}
} ;

struct edges {
	int u, v;
	edges(int u = 0, int v = 0) : u(u), v(v) {}
} e[MAXN];

struct matrix {
	int a[MAXN][MAXN];
	int n;
	matrix() {}
	matrix(int n) : n(n) {
		for (int i = 0; i < n; ++i)
			for (int j = 0; j < n; ++j)
				a[i][j] = 0;
	}
	
	inline matrix operator * (matrix b) {
		matrix ans(n);
		for (int i = 0; i < n; ++i)
			for (int k = 0; k < n; ++k)
				for (int j = 0; j < n; ++j)
					ans.a[i][j] = (ans.a[i][j] + a[i][k] * b.a[k][j]) % mod;
		return ans;
	}
	
	inline matrix operator ^ (int b) {
		matrix ans(n), a = *this;
		for (int i = 0; i < n; ++i) ans.a[i][i] = 1;
		for (; b; b >>= 1, a = a * a)
			if (b & 1) ans = ans * a;
		return ans;
	}
} map, ans;

int method, n, m, step, s, t, x, y, ind;
std::vector<node> edge[MAXN];

inline void add_edge(int x, int y) {
	edge[x].push_back(node(y, ind++));
	edge[y].push_back(node(x, ind++));
}

inline void solve() {
	scanf("%d%d%d%d%d", &n, &m, &step, &s, &t);
	for (int i = 0; i < m; ++i) 
		scanf("%d%d", &x, &y), add_edge(x, y), e[i] = edges(x, y);
	ans = matrix(ind), map = matrix(ind);
	for (int i = 0; i < m; ++i) {
		x = e[i].u, y = e[i].v, ind = (i << 1);
		for (int p = 0; p < edge[y].size(); ++p) {
			node *e = &edge[y][p];
			if (e->id != (ind ^ 1)) map.a[ind][e->id] = 1;
		}
		x = e[i].v, y = e[i].u, ind = (i << 1 | 1);
		for (int p = 0; p < edge[y].size(); ++p) {
			node *e = &edge[y][p];
			if (e->id != (ind ^ 1)) map.a[ind][e->id] = 1;
		}
	}
//	for (int i = 0; i < (m << 1); ++i, std::cout << '\n')
//		for (int j = 0; j < (m << 1); ++j)
//			std::cout << map.a[i][j] << " ";
	for (int p = 0; p < edge[s].size(); ++p) ans.a[0][edge[s][p].id] = 1;
	map = (map ^ step - 1), ans = ans * map;
	for (int p = 0; p < edge[t].size(); ++p) {
		int id = (edge[t][p].id ^ 1);
		method = (method + ans.a[0][id]) % mod;
	}
	printf("%d", method);
}
 
int main() {
	solve();
	return 0;
}